Find wavefunction of harmonic oscillator

[skrat]

Homework Statement

We want to prepare a particle in state $\psi$ under following conditions:
1. Let energy be $E=\frac{5}{4}\hbar \omega$
2. Probability, that we will measure energy greater than $2\hbar \omega$ is $0$
3. $<x>=0$

Homework Equations

The Attempt at a Solution

I have no idea, yet i tried to do some magic.

$E=\frac{5}{4}\hbar \omega$ than I guess that this is actually a sum of all states? or is it not?

So... $\frac{5}{4}\hbar \omega=\sum (n+1/2)\hbar \omega$

The second condition:

if $P(E>2\hbar \omega)=\sum\left | C_n \right |^2E_n=\sum\left | C_n \right |^2\hbar\omega (n+1/2)=0$ than

$P(E<2\hbar \omega)=\sum\left | C_n \right |^2E_n=\sum\left | C_n \right |^2\hbar\omega (n+1/2)=1$

Now I don't know how to use third condition.. :/ I don't even know if this makes any sense?

 

[vanhees71]

Since $5 \hbar \omega/4$ is not an eigenstate of the Hamiltonian,
[tex]\hat{H}=\frac{\hat{p}^2}{2m}+\frac{m \omega^2}{2} \hat{x}^2,[/itex]
there is no eigenstate with that value of the energy. Do you mean, the average energy is $5/4 \hbar \omega$? But then the state is not uniquely defined.

 

[skrat]

Well it only says that the energy is $\frac{5}{4}\hbar \omega$. I also think there is no eigenstate with that value of energy, therefore I assume that this is meant for the average energy.

 

[DrClaude]

$P(E<2\hbar \omega)=\sum\left | C_n \right |^2E_n=\sum\left | C_n \right |^2\hbar\omega (n+1/2)=1$

That should be $P(E \le 2\hbar \omega)$.

Now I don't know how to use third condition.

What is the value of $\langle x\rangle$ for an eigenstate of the h.o.? What happens if you sum two states of the same parity? What happens if you sum two states of opposite parity?

 

[skrat]

I though the value of $<x>$ for any eigenstate of the harmonic oscillator is $0$. But now, when i came across this problem I have to admit that I am not really sure about that.
Thinking in classical physics the average $x$ is $0$.

If i sum the two states of the opposite parity you get nothing. (straight line)

 

[DrClaude]

I though the value of $<x>$ for any eigenstate of the harmonic oscillator is $0$. But now, when i came across this problem I have to admit that I am not really sure about that.
Thinking in classical physics the average $x$ is $0$.

The fact that an eigenstate is either symmetric or anti-symmetric with respect to $x=0$ should clue you in.

If i sum the two states of the opposite parity you get nothing. (straight line)

Really? Try ploting $\psi_0 + \psi_1$.

 

[skrat]

http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/hosc7.html#c1 one could say all in one page...

So, the condition saying that $<x>=0$ tels me that $psi$ is anti-symmetric. Right?

Functions $\psi _0$ and $\psi _1$ are also plotted on that link, however, i can't see where you are going with that?

 

[DrClaude]

So, the condition saying that $<x>=0$ tels me that $psi$ is anti-symmetric. Right?

No. Look carefully at $\psi_0$. What do you expect $\langle x \rangle$ to be?

Functions $\psi _0$ and $\psi _1$ are also plotted on that link, however, i can't see where you are going with that?

We're interested in the sum of wave functions. Here are $\psi_0$, $\psi_1$, and $(\psi_0 + \psi_1)/\sqrt{2}$. What can you say about $\langle x \rangle$ in each case?

 

[skrat]

Aaaha.

$\left | \psi \right |^2$ is the probability to find the particle at x. Clearely, for $\psi _0$ the $<x>=0$. So, for all symmetrical $\psi$ the $<x>$ is $0$. Can we say anything about $<x>$ of anti-symmetrical functions, except that $<x>\neq 0$?

 

[DrClaude]

Sorry, but you're going at it wrong.
$$
\begin{align}
\langle x \rangle &= \int_{-\infty}^{\infty} x | \psi(x) |^2 \; \mathrm{d}x \\
&= \int_{-\infty}^{0} x | \psi(x) |^2 \; \mathrm{d}x + \int_{0}^{\infty} x | \psi(x) |^2 \; \mathrm{d}x \\
&= - \int_{0}^{\infty} (-x) | \psi(-x) |^2 \; \mathrm{d}(-x) + \int_{0}^{\infty} x | \psi(x) |^2 \; \mathrm{d}x \\
&= - \int_{0}^{\infty} x | \psi(x) |^2 \; \mathrm{d}x + \int_{0}^{\infty} x | \psi(x) |^2 \; \mathrm{d}x \\
&= 0
\end{align}
$$
where I made use of $| \psi(-x) |^2 = | \psi(x) |^2$ for $\psi$ either symmetric or anti-symmetric.

If $\psi$ is neither symmetric or anti-symmetric, then $\langle x \rangle$ is most often not zero. So either you make your $\psi$ symmetric or anti-symmetric or (hint, hint) use a clever superposition of symmetric and anti-symmetric wave functions such that $| \psi(x) |^2$ itself is symmetric [STRIKE]or anti-symmetric[/STRIKE].

[Edit: I corrected the silliness just above. $| \psi(x) |^2$ obviously can't be anti-symmetric.]

 

[skrat]

If $\psi$ is neither symmetric or anti-symmetric, then $\langle x \rangle$ is most often not zero. So either you make your $\psi$ symmetric or anti-symmetric or (hint, hint) use a clever superposition of symmetric and anti-symmetric wave functions such that $| \psi(x) |^2$ itself is symmetric or anti-symmetric.

This kinda of frightens me. I guess I don't know a thing about harmonic oscillators or wavefunctions in general.

Since $<x>=0$ in my case, than $\psi$ is a combination of symmetric and anti-symmetric wavefunctions. BUT, how can I be sure that we are talking about $\psi _0$ and $\psi _1$ here and not about $\psi _{3432434}$ and $\psi _{23}$ (random functions)?

 

[DrClaude]

This kinda of frightens me. I guess I don't know a thing about harmonic oscillators or wavefunctions in general.

But you're learning, aren't you?

Since $<x>=0$ in my case, than $\psi$ is a combination of symmetric and anti-symmetric wavefunctions.

I'll repeat: $\psi$ will either be

• a superposition of symmetric wave functions
• a superposition of anti-symmetric wave functions
• a clever superposition of symmetric and anti-symmetric wave functions, such that $| \psi |^2$ is symmetric

BUT, how can I be sure that we are talking about $\psi _0$ and $\psi _1$ here and not about $\psi _{3432434}$ and $\psi _{23}$ (random functions)?

Remember that you are considering here the third criterion for $\psi$. You have the first two criteria that also play a role. I used $\psi_0$ and $\psi_1$ only as an example.

 

[skrat]

But you're learning, aren't you?

I am, but still... I should have known that. It's frustrating me now. -.- Nevermind.

I'll repeat: $\psi$ will either be

• a superposition of symmetric wave functions
• a superposition of anti-symmetric wave functions
• a clever superposition of symmetric and anti-symmetric wave functions, such that $| \psi |^2$ is symmetric

Remember that you are considering here the third criterion for $\psi$. You have the first two criteria that also play a role. I used $\psi_0$ and $\psi_1$ only as an example.

First condition: $E=\frac{5}{4}\hbar \omega$
Second condition: $P(E\leq 2\hbar \omega)=\sum\left | C_n \right |^2E_n=\sum\left | C_n \right |^2\hbar\omega (n+1/2)=1$
and the third is written by you.

If the first condition is not the average energy, than $E=\frac{5}{4}\hbar \omega=\hbar \omega (n+1/2)$. Inserting this in second condition gives me $\sum\left | C_n \right |^2=\frac{4}{5\hbar \omega}$ where $C_n$ can be complex.

Now what? To me this kinda feels like $\psi$ will be a superposition of symmetric and anti-symmetric wave functions.

 

[DrClaude]

Inserting this in second condition gives me $\sum\left | C_n \right |^2=\frac{4}{5\hbar \omega}$ where $C_n$ can be complex.

You have an $\hbar \omega$ too much.

Now what?

What does criterion 2 tell you?

 

[BruceW]

yes. sorry to barge in. But as DrClaude is implying, your current equation for condition 2 is not correct. condition 2 is telling you something about the possible outcome of an energy measurement.

 

[DrClaude]

yes. sorry to barge in. But as DrClaude is implying, your current equation for condition 2 is not correct. condition 2 is telling you something about the possible outcome of an energy measurement.

Thanks Bruce. I read too fast and missed that.

 

[vanhees71]

I think, here a lot is totally confused now. So let's first repeat the basics.

(1) A (pure) state in quantum theory can be described by a wave function $\psi(x)$ that is square integrable, i.e., the integral
$$\int_{\mathbb{R}} \mathrm{d} x |\psi(x)|^2$$
exists.

(2) The probability distribution to find the particle at place $x$ is given by $|\psi(x)|^2$.

(3) Any observable is represented by a self-adjoint operator on the Hilbert space of wave functions.

(4) A possible value of the observable is given by a (generalized) eigenvalue of the self-adjoint operator representing it.

For the harmonic oscillator the Hamiltonian (operator representing the total energy) is given by
$$\hat{H}=\frac{\hat{p}^2}{2m}+\frac{m \omega^2}{2}.$$
The momentum operator is given by $-\mathrm{i} \hbar \frac{\mathrm{d}}{\mathrm{d} x}$. The eigenvalues of $\hat{H}$ are
$$E_n=\frac{\hbar \omega}{2}(2n+1), \quad n \in \{0,1,2,\ldots \}=\mathbb{N}_0.$$

The normalized eigenfunctions of the Hamiltonian $u_n(x)$ build a complete set of orthonormal functions in the Hilbert space, i.e., any state is given by a superposition of these eigenfunctions:
$$\psi(x)=\sum_{n=0}^{\infty} C_n u_n(x), \quad C_n=\langle u_n |\psi \rangle=\int_{\mathbb{R}} \mathrm{d} x \; u_n^*(x) \psi(x).$$
In the following let $\psi$ be normalized, i.e.,
$$\int_{\mathbb{R}} \mathrm{d} x |\psi(x)|^2=\sum_{n=0}^{\infty} |C_n|^2=1.$$

As stated earlier the constraint 1. is impossible to fulfill, because the energy cannot be $5 \hbar \omega/4$. So I guess the statement should be that the mean enery,
$$\langle E \rangle = \sum_{n=0}^{\infty} |C_n|^2 E_n=\frac{5 \hbar \omega}{4}.$$

Now you should be able to put everything together. The trick is to express all the constraints (1)-(3) in terms of the coefficients $C_n$, with (1) corrected as said above (because otherwise the question doesn't make any sense or the answer is trivially that there is no state that fulfills constraint (1)).

 

[skrat]

First condition:

$\sum_{n=0}^{\infty}\left | C_n \right |^2E_n=\frac{5}{4}\hbar \omega$

Second should say that... The energy $E$ of $\psi$ which I am looking is for sure $E\leq 2 \hbar \omega$ so $(n+1/2)\leq 2$ and finally

$n\leq \frac{3}{2}$

So my $\psi$ can only be a combination of $\psi _0$ and $\psi _1$

or not?

First and second condition together now give me

$\left | C_0 \right |^2E_0+\left | C_1 \right |^2E_1=\frac{5}{4}\hbar \omega$

$\left | C_0 \right |^2+3\left | C_1 \right |^2=\frac{5}{2}$

Now I know that $\psi = C_0 \psi _0 + C_1\psi _1$

Third condition, if I understood all of you right, says that

$<x>=\int_{-\infty}^{\infty}\psi ^*x\psi dx=\int_{-\infty}^{\infty}(\left | C_0 \right |^2\left | \psi_0 \right |^2x+\left | C_1 \right |^2\left | \psi_1 \right |^2x)dx=0$

I hope this mens that $\left | C_0 \right |^2+\left | C_1 \right |^2=0$

Now I have a system of two equations with two constants

$\left | C_0 \right |^2+\left | C_1 \right |^2=0$

$\left | C_0 \right |^2+3\left | C_1 \right |^2=\frac{5}{2}$

Which give me $\left | C_1 \right |^2=\frac{5}{4}$ and $\left | C_0 \right |^2=-\frac{5}{4}$ which is a bit confusing since I am working with absolute values here...

What did I do wrong this time?

 

[DrClaude]

First condition:

$\sum_{n=0}^{\infty}\left | C_n \right |^2E_n=\frac{5}{4}\hbar \omega$

Second should say that... The energy $E$ of $\psi$ which I am looking is for sure $E\leq 2 \hbar \omega$ so $(n+1/2)\leq 2$ and finally

$n\leq \frac{3}{2}$

So my $\psi$ can only be a combination of $\psi _0$ and $\psi _1$

Correct.

First and second condition together now give me

$\left | C_0 \right |^2E_0+\left | C_1 \right |^2E_1=\frac{5}{4}\hbar \omega$

$\left | C_0 \right |^2+3\left | C_1 \right |^2=\frac{5}{2}$

Now I know that $\psi = C_0 \psi _0 + C_1\psi _1$

You're fine up to here. Now you should normalize $\psi$ before continuing further.

$<x>=\int_{-\infty}^{\infty}\psi ^*x\psi dx=\int_{-\infty}^{\infty}(\left | C_0 \right |^2\left | \psi_0 \right |^2x+\left | C_1 \right |^2\left | \psi_1 \right |^2x)dx=0$

That is not correct. You can't transform $\psi^* \psi$ into $\left | C_0 \right |^2\left | \psi_0 \right |^2 +\left | C_1 \right |^2\left | \psi_1 \right |^2$.

There are many ways to proceed here. I suggest the following:
Write that integral by correctly substituting $\psi^* \psi$, with coefficients $C_0$ and $C_1$. Then rewrite the $C$'s as $a + i b$, where $a$ and $b$ are real. What you will be looking for inside the integral is a sum of symmetric and anti-symmetric terms. What you want to avoid are terms that have no defined symmetry, as they are the ones that lead to $\langle x \rangle \neq 0$.

 

[vanhees71]

Ok, the average-energy condition is correct, giving indeed
$$|C_0|^2+3 |C_1|^2=\frac{5}{2}.$$
Then what's the condition for the correct normalization of the wave function?

Further, your wave function is
$$\psi_x=C_0 u_0(x)+C_1 u_1(x).$$
The average position is
$$\langle x \rangle = \int_{\mathbb{R}} \mathrm{d} x \psi^*(x) x \psi(x)=\int_{\mathbb{R}} \mathrm{d} x x |\psi(x)|^2.$$
Now carefully (!) calculate $|\psi(x)|^2$ and plug this into the equation for the average. What do you get then?

To finally solve the problem you only need to know that you can choose the overall phase of the wave function arbitrarily, i.e., you can assume, e.g., $C_0 \geq 0$.

 

[skrat]

$|C_0|^2+3 |C_1|^2=\frac{5}{2}$

from normalization also

$|C_0|^2+ |C_1|^2=1$Carefully ( =) ) calculated $\left | \psi (x) \right |^2=\left | C_0 \right |^2\left | \psi _0 \right |^2+C_0^*C_1\psi _0^*\psi _1+C_1^*C_0\psi _1^*\psi _0+\left | C_1 \right |^2\left | \psi _1 \right |^2$

 

[DrClaude]

$\left | \psi (x) \right |^2=\left | C_0 \right |^2\left | \psi _0 \right |^2+C_0^*C_1\psi _0^*\psi _1+C_1^*C_0\psi _1^*\psi _0+\left | C_1 \right |^2\left | \psi _1 \right |^2$

Good. Now, to get $\langle x \rangle = 0$, you need all terms to have a definite parity, which is not the case for $C_0^*C_1\psi _0^*\psi _1$ and $C_1^*C_0\psi _1^*\psi _0$, so you need to get rid of them.

 

[BruceW]

why don't they have definite parity? (playing devil's advocate)

edit: if skrat has notes written down somewhere with the form of $\psi_0$ and $\psi_1$ then it would be possible to explicitly see what $\langle x \rangle$ is, and then work out what $C_0$ and $C_1$ need to be.

 

[skrat]

why don't they have definite parity? (playing devil's advocate)

edit: if skrat has notes written down somewhere with the form of $\psi_0$ and $\psi_1$ then it would be possible to explicitly see what $\langle x \rangle$ is, and then work out what $C_0$ and $C_1$ need to be.

Since I am having some troubles translating what "definite parity" would mean, I am going with the second option.

$\psi _n=\frac{1}{\sqrt{a}\pi ^{1/4}}\sqrt{\frac{1}{2^{n}n!}}H_n(\frac{x}{a})e^{-\frac{x^2}{2a^2}}$

so

$\left \langle x \right \rangle=\int_{-\infty }^{\infty }\psi _1x\psi _0\mathrm{d}x=\frac{1}{a}\frac{1}{\sqrt{\pi }}\int_{-\infty }^{\infty }\frac{1}{\sqrt{2}}(2\frac{x}{a})xe^{-\frac{x^2}{a^2}}=\frac{2}{a^2\sqrt{2\pi }}2\int_{0}^{\infty }x^2e^{-\frac{x^2}{a^2}}=\frac{a}{\sqrt{2}}$

where $\sqrt{\frac{\hbar}{m\omega }}$

I hope everything is ok..

 

[DrClaude]

Since I am having some troubles translating what "definite parity" would mean

I'm sorry if this confuses you. Bruce's method might be easier for you. Just to clarify: I meant that the terms $\psi_1^* x \psi_0$ and $\psi_0^* x \psi_1$ are neither symmetric nor anti-symmetric, and therefore will always contribute something to $\langle x \rangle$, which you figured out from the integral.

So, you need $\langle x \rangle = 0$. What does that set as a condition on $C_0$ and $C_1$?

 

[skrat]

No need to apologize. =)

$C_1C_0^*\frac{a}{\sqrt{2}}+C_1^*C_0\frac{a}{\sqrt{2}}=0$

than... am... $C_1C_0^*+C_1^*C_0=0$.

 

[BruceW]

you got it :)

 

[BruceW]

I'm sorry if this confuses you. Bruce's method might be easier for you. Just to clarify: I meant that the terms $\psi_1^* x \psi_0$ and $\psi_0^* x \psi_1$ are neither symmetric nor anti-symmetric, and therefore will always contribute something to $\langle x \rangle$, which you figured out from the integral.

why are they not symmetric nor anti-symmetric? But anyway, it is true that in this case, they do contribute to $\langle x \rangle$

 

[DrClaude]

No need to apologize. =)

$C_1C_0^*\frac{a}{\sqrt{2}}+C_1^*C_0\frac{a}{\sqrt{2}}=0$

than... am... $C_1C_0^*+C_1^*C_0=0$.

Great. Put that with the other two equations the $C$'s you wrote in #21, and you can get their numerical value (up to an arbitrary phase choice, see vanhees at #20.

 

[DrClaude]

why are they not symmetric nor anti-symmetric?

Sorry, getting tired. I'm not expressing myself correctly here. I was thinking of the total superposition which is neither symmetric nor anti-symmetric if the two cross terms do not add up to zero.

 

[BruceW]

ah right. I understand you now. If we make the total wavefunction $\psi$ either symmetric or anti-symmetric, then we are guaranteed that $\langle x \rangle$ is zero. And we know the mod square of the energy eigenstates are symmetric. Therefore, we can guarantee $\langle x \rangle = 0$ by getting rid of the cross terms.